Re: two integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg79253] Re: two integrals
- From: dimitris <dimmechan at yahoo.com>
- Date: Sun, 22 Jul 2007 04:21:23 -0400 (EDT)
- References: <f7sfpg$ru3$1@smc.vnet.net>
On 21 , 11:23, chuck009 <dmili... at comcast.com> wrote:
> How about just using the built-in CauchyPrinipalValue routine for the numerical check:
>
> In[68]:=
> << NumericalMath`CauchyPrincipalValue`
> In[68]:=
> CauchyPrincipalValue[int2 /. p -> 3/4, {z, 0, {3/4}, 1},
> WorkingPrecision -> 50, PrecisionGoal -> 30]
>
> Out[68]=
> -3.350007071973017439190457717672333631156356729013`31.4358463\
> 52995448
>
>
>
> > On 18 , 09:56, dimitris <dimmec... at yahoo.com>
> > wrote:
> > N[res3 /. p -> 3/4]
>
> > Out[31]=
> > -3.3500070719730175 + 0.*I
>
> > In[34]:=
> > NIntegrate[ArcTan[(1 - z^2)/z^3]*(1/(z - p)) /. p ->
> > 3/4, {z, 0, 3/4 -
> > 10^(-7)}] +
> > NIntegrate[ArcTan[(1 - z^2)/z^3]*(1/(z - p)) /. p
> > p -> 3/4, {z, 3/4 +
> > 10^(-7), 1}]
>
> > Out[34]=
> > -3.350006329606476- -
>
> - -
Of course!
Dimitris